- Argand Diagram

A complex number \(z=a+bi\) can be represented on a diagram called an Argand diagram as

(i) a point with coordinates \((a, b)\)
(ii) a vector


REPRESENTING SUMS AND DIFFERENCES ON ARGAND DIAGRAMS
LESSON 1
Find \(z_1+z_2\) and \(z_1-z_2\) for \(z_1=6+2i\) and \(z_2=4+9i\). Hence, represent \(z_1+z_2\) and \(z_1-z_2\) on Argand diagrams.
SOLUTION
(i) \(z_1+z_2=(6+2i)+(4+9i)=10+11i\)
Letting \(z_3=z_1+z_2\) we see that \(z_3\) is the diagonal of the parallelogram formed with \(z_1\) and \(z_2\) as sides its sides.

(ii) \(z_1-z_2=(6+2i)-(4+9i)=2-7i\)
Letting \(z_3=z_1-z_2\) we see that in the case of the subtraction of complex numbers that \(z_3\) and \(z_2\) represent the sides of the parallelogram with \(z_1\) being one of the diagonals.

THE MODULUS OF A COMPLEX NUMBER
The modulus of a complex number, \(z=a+bi\), is a measure of the magnitude of \(z\), and is written as \(|z|\).
Thus modulus \(z=|z|=\sqrt{a^2+b^2}\).

LESSON 1
Determine the modulus of
(i) \(z_1=1+i\)
(ii) \(z_2=-3+4i\)
(iii) \(z_3=-1-\sqrt{3}i\)
(iv) \(z_4=-5\)
SOLUTION
(i) \(z_1=1+i\)
\(|z_1 |=\sqrt{1^2+1^2}=\sqrt{2}\)

(ii) \(z_2=-3+4i\)
\(|z_2 |=|-3+4i|=\sqrt{(-3)^2+4^2}=5\)

(iii) \(z_3=-1-\sqrt{3} i\)
\(|z_3 |=|-1-\sqrt{3} i|=\sqrt{(-1)^2+(-\sqrt{3})^2}=2\)

(iv) \(z_4=-5\)
\(|z_4 |=5\)

LESSON 2
If \(z_1=-3+4i\) and \(z_2=2-i\), what is \(|z_1-z_2|\)?
SOLUTION
We are trying to find the distance between \(z_1\) and \(z_2\). In other words, what is the distance between the points, say, \(P(-3, 4)\) and \(Q(2, -1)\) on the Argand Diagram?

Since we know the coordinates of the two points we can use the formula for the length of a line segment.
\(|PQ|=\sqrt{(x_2-x_1 )^2+(y_2-y_1 )^2}\)
\(⟹|z_1-z_2 |=\sqrt{(-3-2)^2+(4-(-1))^2}\)
\(=\sqrt{50}\)

THE ARGUMENT OF COMPLEX NUMBER
The angle θ is called the argument of \(z (\arg ⁡z)\) where \(\theta\) is the angle the vector representing the complex number on the Argand diagram makes with the positive real axis. Thus \(\displaystyle \tan⁡θ={b\over a}\). To avoid complications, we use \(–π<θ≤π\) and this is known as the principal argument of \(z\).

LESSON 1
Determine the argument of
(a) \(z_1=1+i\)
(b) \(z_2=-3+4i\)
(c) \(z_3=-1-\sqrt{3} i\)
(d) \(z_4=-5\)
SOLUTION
It is useful to draw a sketch of the complex number so that you know which quadrant the complex number is in and then you can determine the appropriate formula for the argument.
(a) \(z=1+i\)
\(z\) is in QUADRANT 1 since both the real and imaginary parts are positive. Therefore \(\displaystyle \arg ⁡z=\tan^{-1}\left({b\over a}\right)\)
\(\arg⁡ (z_1)=θ\)
\(\displaystyle =\tan^{-1}\left({1\over1}\right)\)
\(\displaystyle ={π\over4}\)

(b) \(z=-3+4i\)
\(z\) is in Quadrant II since the real part is negative and the imaginary part is positive. Therefore, \(\displaystyle \arg ⁡z=π-\tan^{-1}⁡\left({b\over a}\right)\)
\(\arg⁡ (z_2)=θ\)
\(\displaystyle =π-\tan^{-1}\left({4\over3}\right)\)
\(\displaystyle =2.21\)

(c) \(z=-1-\sqrt{3} i\)
Both the imaginary and real parts for \(z\) are negative. This implies that \(z\) is in quadrant IV, \(\displaystyle \arg ⁡z=-π+\tan^{-1}\left({b\over a}\right)\).
\(\arg⁡ (z_3)=θ\)
\(\displaystyle =-π+\tan^{-1}⁡\left({\sqrt{3}\over 1}\right)\)
\(\displaystyle =-{2π\over3}\)
(d) \(z_4=-5\)
This is simply a straight line on the \(x\) – axis.
\(\arg⁡ (z_4)=π\)


MODULUS – ARGUMENT FORM

From the diagram
\(\displaystyle \sin ⁡θ={b\over r}\)
\(⟹r \sin ⁡θ=b\)

\(\displaystyle \cos⁡ θ={a\over r}\)
\(r \cos⁡θ=a\)

Given that
\(z=a+bi\)
replacing \(a\) and \(b\) we get the modulus argument form of a complex number
\(⟹z=r \cos⁡θ+ri \sin⁡θ\)
\(⟹z=r(\cos⁡θ+i \sin⁡θ)\)

LESSON 1
Write the following in modulus – argument form
(a) \(z_1=1+i\)
(b) \(z_2=-3+4i\)
(c) \(z_3=-1-\sqrt{3} i\)
(d) \(z_4=-5\)
SOLUTION
The moduli and arguments for these complex numbers were determined previously. Therefore, the required modulus – argument form is shown directly.
(a) \(z_1=1+i\)
\(\displaystyle z_1=\sqrt{2} \left(\cos⁡ \left({π\over4}\right)+i \sin \left({π\over4}\right)\right)\)

(b) \(z_2=-3+4i\)
\(z_2=5(\cos⁡(2.21)+i \sin⁡(2.21))\)

(c) \(z_3=-1-\sqrt{3} i\)
\(\displaystyle z_3=2\left(\cos\left(-{2π\over3}\right)+i\sin\left(-{2π\over3}\right)\right)\)

(d) \(z_4=-5\)
\(z_4=5(\cos ⁡π+i \sin ⁡π)\)

LESSON 2
Given that \(z_1=1+\sqrt{3} i\) and \(z_2=1-i\), determine (i) \(z_1 z_2\) and (ii) \(\displaystyle {z_1\over z_2}\) in modulus – argument form.
SOLUTION
  • Determine the modulus and principal argument of \(z_1\)
\(|z_1 |=\sqrt{1^2+(\sqrt{3})^2}=2\)
\(\displaystyle \arg⁡ (z_1)=\tan^{-1}\left({\sqrt{3}\over1}\right)={π\over3}\)
  • Determine the modulus and principal argument of \(z_2\)
\(|z_2 |=\sqrt{1^2+1^2}=\sqrt{2}\)
\(\displaystyle \arg⁡ (z_2)=-\tan^{-1}⁡\left({1\over1}\right)={π\over4}\)
  • Use the appropriate formulae to determine the required solutions.
\(z_1 z_2=r_1 r_2 [\cos⁡(A+B)+i \sin⁡(A+B)]\)
\(\displaystyle z_1 z_2=2\sqrt{2} \left[\cos\left({π\over3}+{π\over4}\right)+i \sin\left({π\over3}+{π\over4}\right) \right]\)
\(\displaystyle =2\sqrt{2} \left[\cos⁡\left({7π\over12}\right)+i \sin⁡\left({7π\over12}\right)\right]\)

\(\displaystyle {z_1\over z_2}={r_1\over r_2}\left[\cos⁡(A-B)+i \sin⁡(A-B)\right]\)
\(\displaystyle {z_1\over z_2} ={2\over \sqrt{2}} \left[\cos\left({π\over3}-{π\over4}\right)+i \sin\left({π\over3}-{π\over4}\right)\right]\)
\(\displaystyle =\sqrt{2} \left[\cos⁡ \left({π\over12}\right)+i \sin⁡\left({π\over12}\right)\right]\)